, 2008

Let $H$ be the orthocenter of $\triangle ABC$, lines $AA_1$ and $BC$ intersect at point $A_3$. Then $\triangle A_3BA_1 \sim \triangle ABA_3$, $\triangle A_1A_3C \sim \triangle ACA_3 \Rightarrow CA_3^2 = A_3A_1 \cdot AA_3 = BA_3^2 \Rightarrow CA_3 = BA_3$ (fig. 6).

Since $\angle BA_1C = 180^\circ - \angle BAC$, by means of the triangles $BB_0A$ and $CC_0A$ we can easily find that $\angle BHC = 180^\circ - \angle BAC$, i.e. $\angle BA_1C = \angle BHC$. Therefore,

Fig. 6

$B, H, A_1, C$ are circular points and $\angle HA_1C = 180^\circ - \angle HBC = 90^\circ + \gamma$. Since $\angle A_3A_1C = \angle A_1AC + \angle A_1CA = \gamma \Rightarrow \angle HA_1A_3 = \angle HA_1C - \angle A_3A_1C = 90^\circ - \gamma + \gamma = 90^\circ = \angle HA_1A$. Hence $A, C_0, H, A_1$ are circular points. Therefore, $\angle AA_1C_0 = \angle AHC_0 = 180^\circ - \angle AHC = 180^\circ - A_0HC_0 = \beta$ since points $B, C_0, H, A_1$ are also circular. Therefore, $\triangle AA_1C_0 \sim \triangle ABA_3$.

If we find point $C_3 = CC_1 \cap AB$ similarly to the previous constructions, it will be a midpoint of the side $AB$. Therefore, $C_3$ and $A_2$ are midpoints of the respective sides of the similar triangles. Thus, $\angle C_0A_2A_1 = \angle A_3C_3B = \alpha$. But $\angle C_0A_0A_3 = 90^\circ + \angle AA_0C_0 = 90^\circ + \angle C_0BH = 180^\circ - \alpha$ and therefore $A_0, C_0, A_2, A_3$ are circular points. This implies that point $A_2$ belongs to the circle of nine points of $\triangle ABC$. Similarly points $B_2$ and $C_2$ also belong to this circle.

Then we obtain that $\angle C_0A_1A_2 = \beta$, $\angle C_0A_2A_1 = \alpha$, therefore $\triangle A_1A_2C_0 \sim \triangle ABC \Rightarrow \frac{C_0A_2}{AC} = \frac{A_2A_1}{AB}$. Similarly $\frac{B_0A_2}{AB} = \frac{A_2A_1}{AC} \Rightarrow \frac{C_0A_2}{B_0A_2} = \frac{AC^2}{AB^2} = \frac{\sin \angle C_0A_0A_2}{\sin \angle B_0A_0A_2}$ since all the points $A_0, B_0, C_0,$ and $A_2$ are on one circle. Similarly $\frac{B_0C_2}{A_0C_2} = \frac{BC^2}{AC^2} = \frac{\sin \angle B_0C_0C_2}{\sin \angle A_0C_0C_2}$ and $\frac{A_0B_2}{C_0B_2} = \frac{AB^2}{CB^2} = \frac{\sin \angle A_0B_0B_2}{\sin \angle C_0B_0B_2}$.

Let's multiply the last equalities $\frac{\sin \angle C_0A_0A_2}{\sin \angle B_0A_0A_2} \cdot \frac{\sin \angle B_0C_0C_2}{\sin \angle A_0C_0C_2} \cdot \frac{\sin \angle A_0B_0B_2}{\sin \angle C_0B_0B_2} = 1$. The Ceva theorem implies that lines $A_0A_2$, $B_0B_2$, and $C_0C_2$ intersect at one point.